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Difference between revisions of "2005 AMC 12A Problems/Problem 13"

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== Problem ==
 
== Problem ==
 
In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and <math>E</math> are replaced by the
 
In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and <math>E</math> are replaced by the
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numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
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numbers <math>3, 5, 6, 7</math> and <math>9,</math> although not necessarily in that order. The sums of the
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numbers at the ends of the line segments <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{CE}</math>, and <math>\overline{EA}</math> form an
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numbers at the ends of the line segments <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{DE}</math>, and <math>\overline{EA}</math> form an
 
arithmetic sequence, although not necessarily in that order. What is the middle
 
arithmetic sequence, although not necessarily in that order. What is the middle
 
term of the arithmetic sequence?
 
term of the arithmetic sequence?
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</math>
 
</math>
  
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== Solution ==
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== Solutions ==
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'''
 
  
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===Solution 1===
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<math>(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)</math> (i.e., each number is counted twice). The sum <math>A + B + C + D + E</math> will always be <math>3 + 5 + 6 + 7 + 9 = 30</math>, so the arithmetic sequence has a sum of <math>2 \cdot 30 = 60</math>. The middle term must be the average of the five numbers, which is <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</math>.
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===Solution 2===
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Let the terms in the arithmetic sequence be <math>a</math>, <math>a + d</math>, <math>a + 2d</math>, <math>a + 3d</math>, and <math>a + 4d</math>. We seek the middle term <math>a + 2d</math>.
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These five terms are <math>A + B</math>, <math>B + C</math>, <math>C + D</math>, <math>D + E</math>, and <math>E + A</math>, in some order. The numbers <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are equal to 3, 5, 6, 7, and 9, in some order, so
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<cmath>A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.</cmath>
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Hence, the sum of the five terms is
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<cmath>(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.</cmath>
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But adding all five numbers, we also get <math>a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d</math>, so
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<cmath>5a + 10d = 60.</cmath>
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Dividing both sides by 5, we get <math>a + 2d = \boxed{12}</math>, which is the middle term. The answer is (D).
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===Solution 3===
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Not too bad with some logic and the awesome guess and check. Let <math>A=6</math>. Then let <math>B=7,E=5</math> and <math>C=3,D=9</math>. Our arithmetic sequence is <math>10,11,12,13,14</math> so our answer is <math>12 \Longrightarrow \mathrm{(D)}</math>.
  
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<math>(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)</math> (i.e., each number is counted twice). The sum <math>A + B + C + D + E</math> will always be <math>3 + 5 + 6 + 7 + 9 = 30</math>, so the arithmetic sequence has a sum of <math>2 \cdot 30 = 60</math>. The middle term must be the average of the five numbers, which is <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</math>.
+
Solution by franzliszt
  
 
== See also ==
 
== See also ==

Latest revision as of 10:01, 22 August 2024

Problem

In the five-sided star shown, the letters $A$, $B$, $C$, $D$ and $E$ are replaced by the numbers $3, 5, 6, 7$ and $9,$ although not necessarily in that order. The sums of the numbers at the ends of the line segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, and $\overline{EA}$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

[asy] draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); label("$A$",(0.5,1.54),N); label("$B$",(1,0),SE); label("$C$",(-0.31,0.95),W); label("$D$",(1.31,0.95),E); label("$E$",(0,0),SW); [/asy]

$(\mathrm {A}) \ 9 \qquad (\mathrm {B}) \ 10 \qquad (\mathrm {C})\ 11 \qquad (\mathrm {D}) \ 12 \qquad (\mathrm {E})\ 13$

Solutions

Solution 1

$(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)$ (i.e., each number is counted twice). The sum $A + B + C + D + E$ will always be $3 + 5 + 6 + 7 + 9 = 30$, so the arithmetic sequence has a sum of $2 \cdot 30 = 60$. The middle term must be the average of the five numbers, which is $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$.

Solution 2

Let the terms in the arithmetic sequence be $a$, $a + d$, $a + 2d$, $a + 3d$, and $a + 4d$. We seek the middle term $a + 2d$.

These five terms are $A + B$, $B + C$, $C + D$, $D + E$, and $E + A$, in some order. The numbers $A$, $B$, $C$, $D$, and $E$ are equal to 3, 5, 6, 7, and 9, in some order, so \[A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.\] Hence, the sum of the five terms is \[(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.\] But adding all five numbers, we also get $a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d$, so \[5a + 10d = 60.\] Dividing both sides by 5, we get $a + 2d = \boxed{12}$, which is the middle term. The answer is (D).

Solution 3

Not too bad with some logic and the awesome guess and check. Let $A=6$. Then let $B=7,E=5$ and $C=3,D=9$. Our arithmetic sequence is $10,11,12,13,14$ so our answer is $12 \Longrightarrow \mathrm{(D)}$.

Solution by franzliszt

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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